21. A solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on the inclined plane, then

1. 1. solid sphere reaches the bottom first

   
   
   

   2. solid sphere reaches the bottom last

   
   
   

   3. disc will reach the bottom first

   
   
   

   4. all reach the bottom at the same time

   
   
   

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1. View Hint View Answer Discuss in Forum

   For solid sphere,	K²	=	2
   	R²		5

   
   

   For disc and solid cylinder,	K²	=	1
   	R²		2

   
   As K²/R² for solid sphere is smallest, it takes minimum time to reach the bottom of the incline 

   Correct Option: A

   For solid sphere,	K²	=	2
   	R²		5

   
   

   For disc and solid cylinder,	K²	=	1
   	R²		2

   
   As K²/R² for solid sphere is smallest, it takes minimum time to reach the bottom of the incline 

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22. A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated  with rotation is

1. 1. 2/5

   
   
   

   2. 2/7

   
   
   

   3. 3/5

   
   
   

   4. 3/7

   
   
   

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1. View Hint View Answer Discuss in Forum

   		=	1	MK²ω²
   	Kr		2
   	E		1	Mω²[K² + R²]
   			y²

   
   

   =	K²
   	K² + R²

   
   

   =	2/5	=	2
   	1 + 2/5		7

   
   

   Here, K² =	2	R²
   	5

   Correct Option: B

   		=	1	MK²ω²
   	Kr		2
   	E		1	Mω²[K² + R²]
   			y²

   
   

   =	K²
   	K² + R²

   
   

   =	2/5	=	2
   	1 + 2/5		7

   
   

   Here, K² =	2	R²
   	5

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23. A solid cylinder and a hollow cylinder both of the same mass and same external diameter are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

1. 1. Both together

   
   
   

   2. Solid cylinder

   
   
   

   3. One with higher density

   
   
   

   4. Hollow cylinder

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Torque, Iα = ƒ. R.
   Using Newton's IInd law, mg sin θ – ƒ = ma
   ∵ pure rolling is there, a = Rα
   

   ⇒ mg sin θ -	Iα	= ma
   	R

   
   

   ⇒ mg sin θ -	Ia	= ma
   	R²

   
   

   	∵ α =	a		R
   		αβ

   
   

   or, acceleration, a =	mg sin θ
   	I/R² + m

   
   

   Using, s = ut +	1	at²
   	2

   
   

   or, s =	1	at² ⇒ t α	1
   	2		√a

   
   t minimum means a should be more. This is possible when I is minimum which is the case for solid cylinder.
   Therefore, solid cylinder will reach the bottom first.

   Correct Option: B

   
   Torque, Iα = ƒ. R.
   Using Newton's IInd law, mg sin θ – ƒ = ma
   ∵ pure rolling is there, a = Rα
   

   ⇒ mg sin θ -	Iα	= ma
   	R

   
   

   ⇒ mg sin θ -	Ia	= ma
   	R²

   
   

   	∵ α =	a		R
   		αβ

   
   

   or, acceleration, a =	mg sin θ
   	I/R² + m

   
   

   Using, s = ut +	1	at²
   	2

   
   

   or, s =	1	at² ⇒ t α	1
   	2		√a

   
   t minimum means a should be more. This is possible when I is minimum which is the case for solid cylinder.
   Therefore, solid cylinder will reach the bottom first.

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24. The moment of inertia of a body about a given axis is 1.2 kg m². Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of 25  radian/sec² must be applied about that axis for a duration of

1. 1. 4 seconds

   
   
   

   2. 2 seconds

   
   
   

   3. 8 seconds

   
   
   

   4. 10 seconds

   
   
   

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1. View Hint View Answer Discuss in Forum

   I = 1.2  kg m², E_r = 1500 J,
   α = 25 rad/sec², ω₁ = 0, t  = ?
   

   As Er =	1	Iω², ω = √	2Er
   	2		I

   
   

   = √	2 × 1500	= 50 rad/sec
   	1.2

   
   From ω₂ = ω₁ + αt
   50 = 0 + 25 t, t = 2 seconds

   Correct Option: B

   I = 1.2  kg m², E_r = 1500 J,
   α = 25 rad/sec², ω₁ = 0, t  = ?
   

   As Er =	1	Iω², ω = √	2Er
   	2		I

   
   

   = √	2 × 1500	= 50 rad/sec
   	1.2

   
   From ω₂ = ω₁ + αt
   50 = 0 + 25 t, t = 2 seconds

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25. Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis ⊥ to its plane and passing through a point on its rim will be

1. 1. 5 I

   
   
   

   2. 3 I

   
   
   

   3. 6 I

   
   
   

   4. 4 I

   
   
   

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1. View Hint View Answer Discuss in Forum

   M.I of uniform circular disc about its diameter = I
   According to theorem of perpendiclar axes,
   

   M.I. of disc about its axis =	1	MR² = 2I
   	2

   
   

   Applying theorem of  | | axes,		∵ I =	1	mr²
   			4

   
   M.I of disc about the given axis
   = 2I + mr² = 2I + 4I = 6I

   Correct Option: C

   M.I of uniform circular disc about its diameter = I
   According to theorem of perpendiclar axes,
   

   M.I. of disc about its axis =	1	MR² = 2I
   	2

   
   

   Applying theorem of  | | axes,		∵ I =	1	mr²
   			4

   
   M.I of disc about the given axis
   = 2I + mr² = 2I + 4I = 6I

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